a, Ta co pthh
2Na + 2H2O \(\rightarrow\) 2NaOH + H2
Theo de bai ta co
nNa=\(\dfrac{4,6}{23}=0,2mol\)
b, Theo pthh
nNaOH=nNa=0,2 mol
\(\Rightarrow\) khoi luong cua NaOH thu duoc la
mNaOH=0,2.40=8g
c, Theo pthh
nH2=\(\dfrac{1}{2}nNa=\dfrac{1}{2}.0,2=0,1mol\)
\(\Rightarrow\)VH2=0,1.22,4=2,24 l